Factorial Linear Forms and Denominators: Detailed Proofs and Rationality Criteria

Value sets do not determine the weighted residue sum.

Garaev, Luca and Shparlinski’s harmonic-sum estimate [20] concerns unconditioned harmonic sums, not sums restricted to the indices where $n!\equiv 1(\mathrm{mod}q)$. The value-set literature addresses different invariants. Banks, Luca, Shparlinski and Stichtenoth [25] provide historical missing-value context; Klurman and Munsch [26] distinguish unconditional averages from GRH-dependent improvements. Grebennikov, Sagdeev, Semchankau and Vasilevskii [27] study sizes of value and quotient sets, as do the factorial-residue and representation results of Hu [22] and Garaev and Pardo [23]. None is a noncancellation theorem for the reciprocal unit weights here. A lower bound for the number of distinct residues does give an upper bound on the multiplicity of residue $1$, since every other value needs at least one index. But that bound does not determine the reciprocal unit weights or test whether they cancel modulo $q$. The two explicit cancellations in §2 illustrate why the weights matter. Hu’s September preprint [24] considers pairwise distinct maps $T_{\theta }(x)=b_{\theta }+c_{\theta }/x$ over ${\mathbb{F}}_p$, with $c_{\theta }\ne 0$. An internal transition is a pair $(x,\theta )$ for which $x\in A\setminus \{0\}$ and $T_{\theta }(x)\in A$. With at least $M$ distinct such pairs, at most $|A|$ parameters, and a bound $\mu \ge 1$ on the number of ordered parameter pairs representing each nonidentity map $T_{\eta }{T}_{\theta }^{-1}$, the theorem gives $|A|\gg min\{M,p{\}}^{8/15}{\mu }^{-4/15}$. Its incidence estimate comes from Stevens and de Zeeuw [28]. To apply this to the residue classes of indices satisfying $n!\equiv 1(\mathrm{mod}q)$, one would first have to specify the set $A$ and the transition maps. The number of such indices is not automatically a lower bound for $M$: different indices may produce the same pair $(x,\theta )$, and their factorial values are all $1$. One must prove that enough distinct transitions remain, as well as bound the ordered-pair multiplicity $\mu$. The unconditioned value-set theorem does not provide these additional facts.

The shared denominator and the gap must be controlled together.

For the blocks selected from first prime occurrences, the second inequality in (27) is

The first inequality is (25). Both must hold at the same arbitrarily large parameters.

The order of quantifiers is part of the proposed result: for every bound, one parameter must satisfy both inequalities. Two unrelated infinite subsequences would not prove the combined criterion.

For a prime $r$ and an integer $e\ge 1$, let $h_{r,e}(p)=\mathrm{\#}\{i\in I_p:r^e\mid i!-1\}$. The exponent of $r$ in the shared part is at least $e$ exactly when $h_{r,e}(p)>1$. Removing the exponent already present in $F_p$ therefore gives

When $r\nmid F_p$, this counts the positive exponents for which two or more denominators are divisible by $r^e$. The spacing bound $h_{r,e}(p)(e+1)\le 2p+e-2$ follows from the gap-product argument: if $i<j$ are two such indices, then $j<r$ and $0<r^e\le j!/i!-1<r^{j-i}$, so $j-i\ge e+1$. All indices lie between $2$ and $2p-1$, giving the bound when $h_{r,e}(p)>0$. It also holds when $h_{r,e}(p)=0$. To estimate $\log {\tilde{C}}_p$, these exponent counts must be multiplied by $\log r$ and summed over primes. A count ignoring the sizes of the primes does not give (29). For a prime $q$, Wilson reflection of an odd index $n<q$ contributes a repeated divisor only when the reflected index is distinct and in the same block, with both indices at least $2$. The endpoint $n=q-2$ reflects to $1$ and contributes no such pair, as shown in §2.

Even for two admissible indices, the reflection identity is only modulo $q$ [8]. For example, exact multiplication gives

where $361=971-609-1$. Both indices lie in $I_{306}$, but their factorial denominators have $971$-exponents at least $2$ and exactly $1$, respectively. Thus both indices satisfy the congruence modulo $q$, but need not satisfy it modulo $q^e$ for $e>1$. Each level in the valuation sum requires its own congruence; neither distinctness nor a higher exponent may be inferred from the prime-level identity alone. Theorem 12 of Garaev, Luca and Shparlinski [7] is a multiplicity bound and supplies no lower bound for the distance to the next integer in (25).

For the first inequality alone, a joint mean bound

on nonempty sets $I_X\subseteq [X,2X]\cap \mathbb{N}$, with $R_p>1$, a fixed $0<\epsilon <1$, and arbitrarily large $X$, would suffice: at least one summand is at most the average, while the required upper bound is $\log (K_p/((2p+1)F_p))=p\log p+O(p)$, uniformly for $p\in [X,2X]$. Two separate estimates on unrelated sets of indices would not give this joint comparison.

What two gap-product tests do not show.

For integers $h\ge 1$ and $x\ge 0$, let $F_h(X)=\prod _{j=1}^h(X+j)$. Suppose $x+h<q$ for a prime $q$. If $x!$ and $(x+h)!$ both equal $1$ modulo $q$ and the harmonic sums through these indices are equal modulo $q$, then $F_h(x)=1$ and $F_h^{\prime }(x)=F_h(x)\sum _{j=1}^h(x+j{)}^{-1}=0$ in that field. An exact calculation gives

Thus a uniform bound of one repeated root is false. But $4626!\equiv 442$ and $7848!\equiv 6300$ modulo $12487$, so neither root also satisfies $x!\equiv 1$. Counting repeated roots of $F_h-1$ therefore does not count the indices required by the factorial problem. These are finite modular calculations, not a bound on the number of indices satisfying both congruences.

Stewart’s gap-product estimate [8] requires the two products to be unequal. He supplies that step separately: in the proof of his bound (14), a prime-distribution argument separates the products when the common divisor is large; the smaller-divisor case is immediate [8].

The non-power theorem gives a different argument when the gaps are adjacent. For integers $2\le i<j<k$, equality $j!/i!=k!/j!$ would make $k!/i!$ a square, contrary to that theorem [31]. Selecting two short gaps with nearly equal lengths need not preserve adjacency, so this observation alone cannot replace Stewart’s nonvanishing argument. The wider equal-product problem has a separate history [29]; the doubled-length theorem discussed by Nguyen Xuan Tho [30] is another special case, not arbitrary near-equal-length nonvanishing. No stronger lcm exponent is inferred from these local observations.

Prime powers already seen at earlier indices.

Recall that ${\mathrm{\Delta }}_n=P_n^{\mathrm{\Delta }}/Q_n^{\mathrm{\Delta }}$ is the positive gap in lowest terms. Define

Thus the product includes the full power of a prime in $n!-1$ when that prime has appeared before but has smaller exponent in $Q_n^{\mathrm{\Delta }}$. The recurrence for the gaps is

For a prime $r$ in the product, put $e=v_r(n!-1)$. The reduced denominator of $n{\mathrm{\Delta }}_n-b_n$ has $r$-exponent at most $v_r(Q_n^{\mathrm{\Delta }})<e$, whereas $1/(n!-1)$ has exponent $e$. Clearing denominators leaves exactly one numerator term nonzero modulo $r$, so the reduced denominator of ${\mathrm{\Delta }}_{n+1}$ has exponent exactly $e$. Hence ${\mathcal{A}}_n\mid Q_{n+1}^{\mathrm{\Delta }}$ and, when ${\mathcal{A}}_n>1$, $P_{n+1}^{\mathrm{\Delta }}\not\equiv 0(\mathrm{mod}{\mathcal{A}}_n)$. The earlier occurrence of $r$ is not used in this divisibility argument; it restricts the product to the primes proposed for this approach. A sufficient quantitative estimate, at arbitrarily large indices $n$, with $m=n+1$ and $d={\mathcal{A}}_n>1$, is

It implies the upper alternative of (21). A nonzero least representative can equal one even when the modulus is large. Neither a lower bound for $\log {\mathcal{A}}_n$ nor infinitely many increases in a previously occurring prime’s exponent therefore proves (30): the least representative itself must satisfy the displayed inequality. Repeated prime squares can first occur before the Wilson index $r-2$: for example, $971$ divides $361!-1$ and its square first divides $609!-1$, with $609<971-2$. Such an example rules out confinement to that last index but supplies no unbounded family satisfying the required inequality.

The upper alternative in the finite test.

It would suffice for (21) to hold at arbitrarily large indices; an unbounded set of prime indices would suffice as well. The upper alternative is exactly $((m+2)m!-2)Q_m^{\mathrm{\Delta }}\le m^2(m!-1)P_m^{\mathrm{\Delta }}$. Replacing $P_m^{\mathrm{\Delta }}$ by its least nonnegative residue modulo a specified divisor of $Q_m^{\mathrm{\Delta }}$ gives a stronger sufficient test. No bound forcing this inequality at arbitrarily large indices is established here. The full recurrence for the actual scaled partial sums, including its initial value, is fixed; a rational example for a weakened recurrence does not satisfy that full specification (§B.7). Nor does a zero canonical digit assert membership in the specified real interval.

Conditions at twice a prime.

For an odd prime $p$, the carry range and $Z_{2p}=2p{Z}_{2p-1}+1-b_{2p}$ give

Reduction modulo $p$ forces $b_{2p}\equiv 1(\mathrm{mod}p)$, and the two displayed values are the only possibilities in $[-1,2p-1]$; division by $p$ gives the predecessor conditions. For rational $S=a/q$ and $p>q$, the tail estimate gives $Z_{2p}=(2p)!S$, which is divisible by $p^2$. A proof through $p^2\nmid Z_{2p}$ must therefore exclude both displayed alternatives at the same prime, for arbitrarily large primes. That is a requirement of this particular test, not of every irrationality argument. Rationality also forces $b_{2p}=1$ and $Z_{2p-1}=(2p-1)!a/q$, with $p\mid Z_{2p-1}$: here $q\mid (p-1)!$, and the remaining factorial factors include $p$. Thus either $b_{2p}\ne 1$ at arbitrarily large odd primes or $p\nmid Z_{2p-1}$ at arbitrarily large odd primes would already contradict rationality. The first is a special case of Theorem 4.1; the second uses the same factorial-clearing argument. Neither unbounded occurrence is proved here.

Criteria from the literature that do not apply.

Duverney’s Theorem 3.1 assumes, among other conditions, quadratic growth $c{u}_n^2\le u_{n+1}\le c^{\prime }{u}_n^2$ for positive constants $c,c^{\prime }$. For $u_n=n!-1$, however, $u_{n+1}/u_n^2\to 0$. His Corollary 3.2 assumes convergence of the signed series $\sum _n(u_{n+1}/u_n^2-1)$ in (3.6). Here its terms tend to $-1$, so that hypothesis fails as well [10]. For a single denominator sequence the rapid-growth criterion goes back to Erdős [3]; Barreto, Kang, Kim, Kovač and Zhang treat products of consecutive denominators and weighted extensions [12]. For $a_n=n!-1$, $\log (n!-1)=O(n\log n)$ makes $(n!-1{)}^{1/{\psi }^n}\to 1$ for every fixed $\psi >1$, so neither Erdős’s hypothesis $\underset{n}{lim sup}{a}_n^{1/2^n}=\mathrm{\infty }$ nor the growth hypotheses of their Theorems 2 and 3 hold. These criteria do not decide $S$; the useful content is the target these papers identify: a subsequence with clearing integers small enough for the corresponding remainders, or enough exact cancellation in those integers. Theorem 3.4 measures the cost of clearing all summands before reduction; it does not bound cancellation in their sum. Theorem 2.1 tests whether a maximal prime power survives reduction. Neither supplies the upper bound for a clearing integer, relative to its remainder, that these arguments need.

No irrationality conclusion is obtained here. The finite conclusion is that every rational representation $S=a/q$, $q>0$, must satisfy both $q\nmid 299999!$ and $q\ge 2^{39990}>{10}^{12038}$. The criteria above require a condition at arbitrarily large indices. Extending either finite computation alone does not establish it.